Calculus

Fourier Series

Definition of Fourier Series and Typical Examples

Page 1
Problems 1-2
Page 2
Problems 3-6

Baron Jean Baptiste Joseph Fourier \(\left( 1768-1830 \right) \) introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related.

To consider this idea in more detail, we should introduce some definitions and common terms.

Baron Jean Baptiste Joseph Fourier  (1768−1830)

Basic Definitions

A function \(f\left( x \right)\) is said to have period \(P\) if \(f\left( {x + P} \right) = f\left( x \right)\) for all \(x.\) Let the function \(f\left( x \right)\) has period \(2\pi.\) In this case, it is enough to consider behavior of the function on the interval \(\left[ { – \pi ,\pi } \right].\)

  1. Suppose that the function \(f\left( x \right)\) with period \(2\pi\) is absolutely integrable on \(\left[ { – \pi ,\pi } \right]\) so that the following so-called Dirichlet integral is finite:
    \[\int\limits_{ – \pi }^\pi {\left| {f\left( x \right)} \right|dx} \lt \infty ;\]
  2. Suppose also that the function \(f\left( x \right)\) is a single valued, piecewise continuous (must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima).

If the conditions \(1\) and \(2\) are satisfied, the Fourier series for the function \(f\left( x \right)\) exists and converges to the given function (see also the Convergence of Fourier Series page about convergence conditions.)

At a discontinuity \({x_0}\), the Fouries Series converges to
\[\lim\limits_{\varepsilon \to 0} \frac{1}{2}\left[ {f\left( {{x_0} – \varepsilon } \right) – f\left( {{x_0} + \varepsilon } \right)} \right].\] The Fourier series of the function \(f\left( x \right)\) is given by
\[{f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}} ,}\] where the Fourier coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}\) are defined by the integrals
\[
{{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} ,\;\;\;}\kern-0.3pt
{{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nx dx} ,\;\;\;}\kern-0.3pt
{{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nx dx} .}
\] Sometimes alternative forms of the Fourier series are used. Replacing \({{a_n}}\) and \({{b_n}}\) by the new variables \({{d_n}}\) and \({{\varphi_n}}\) or \({{d_n}}\) and \({{\theta_n}},\) where
\[
{{d_n} = \sqrt {a_n^2 + b_n^2} ,\;\;\;}\kern-0.3pt
{\tan {\varphi _n} = \frac{{{a_n}}}{{{b_n}}},\;\;\;}\kern-0.3pt
{\tan {\theta _n} = \frac{{{b_n}}}{{{a_n}}},}
\] we can write:
\[
{f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{d_n}\sin \left( {nx + {\varphi _n}} \right)} \;\;}\kern-0.3pt{\text{or}\;\;}
{f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{d_n}\cos\left( {nx + {\theta _n}} \right)} .}
\]

Fourier Series of Even and Odd Functions

The Fourier series expansion of an even function \(f\left( x \right)\) with the period of \(2\pi\) does not involve the terms with sines and has the form:
\[{f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{a_n}\cos nx} ,}\] where the Fourier coefficients are given by the formulas
\[
{{a_0} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)dx} ,\;\;\;}\kern-0.3pt
{{a_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\cos nxdx} .}
\] Accordingly, the Fourier series expansion of an odd \(2\pi\)-periodic function \(f\left( x \right)\) consists of sine terms only and has the form:
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {{b_n}\sin nx} ,\] where the coefficients \({{b_n}}\) are
\[{b_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin nxdx} .\] Below we consider expansions of \(2\pi\)-periodic functions into their Fourier series, supposing that these expansions exist and are convergent.

Solved Problems

Click on problem description to see solution.

 Example 1

Let the function \(f\left( x \right)\) be \(2\pi\)-periodic and suppose that it is presented by the Fourier series:
\[{f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}}}\] Calculate the coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}.\)

 Example 2

Find the Fourier series for the square \(2\pi\)-periodic wave defined on the interval \(\left[ { – \pi ,\pi } \right]:\)
\[
{f\left( x \right) \text{ = }}\kern0pt
{\begin{cases}
0, & \text{if} & – \pi \le x \le 0 \\
1, & \text{if} & 0 < x \le \pi
\end{cases}.}
\]

 Example 3

Find the Fourier series for the sawtooth wave defined on the interval \(\left[ { – \pi ,\pi } \right]\) and having period \(2\pi.\)

 Example 4

Let \(f\left( x \right)\) be a \(2\pi\)-periodic function such that \(f\left( x \right) = {x^2}\) for \(x \in \left[ { – \pi ,\pi } \right].\) Find the Fourier series for the parabolic wave.

 Example 5

Find the Fourier series for the triangle wave
\[
{f\left( x \right) \text{ = }}\kern0pt
{\begin{cases}
\frac{\pi }{2} + x, & \text{if} & – \pi \le x \le 0 \\
\frac{\pi }{2} – x, & \text{if} & 0 \lt x \le \pi
\end{cases},}
\] defined on the interval \(\left[ { – \pi ,\pi } \right].\)

 Example 6

Find the Fourier series for the function
\[
{f\left( x \right) \text{ = }}\kern0pt
{\begin{cases}
-1, & \text{if} & – \pi \le x \le – \frac{\pi }{2} \\
0, & \text{if} & – \frac{\pi }{2} \lt x \le \frac{\pi }{2} \\
1, & \text{if} & \frac{\pi }{2} \lt x \le \pi
\end{cases},}
\] defined on the interval \(\left[ { – \pi ,\pi } \right].\)

Example 1.

Let the function \(f\left( x \right)\) be \(2\pi\)-periodic and suppose that it is presented by the Fourier series:
\[{f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}}}\] Calculate the coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}.\)

Solution.

To define \({{a_0}},\) we integrate the Fourier series on the interval \(\left[ { – \pi ,\pi } \right]:\)
\[
{\int\limits_{ – \pi }^\pi {f\left( x \right)dx} }
= {\pi {a_0} }+{ \sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ – \pi }^\pi {\cos nxdx} }\right.}+{\left.{ {b_n}\int\limits_{ – \pi }^\pi {\sin nxdx} } \right]}}
\] For all \(n \gt 0\),
\[
{{\int\limits_{ – \pi }^\pi {\cos nxdx} }={ \left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{ – \pi }^\pi }={ 0\;\;}{\text{and}\;\;\;}}\kern-0.3pt
{{\int\limits_{ – \pi }^\pi {\sin nxdx} }={ \left. {\left( { – \frac{{\cos nx}}{n}} \right)} \right|_{ – \pi }^\pi }={ 0.}}
\] Therefore, all the terms on the right of the summation sign are zero, so we obtain
\[
{\int\limits_{ – \pi }^\pi {f\left( x \right)dx} = \pi {a_0}\;\;\text{or}\;\;\;}\kern-0.3pt
{{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} .}
\] In order to find the coefficients \({{a_n}},\) we multiply both sides of the Fourier series by \(\cos mx\) and integrate term by term:
\[
{\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} }
= {\frac{{{a_0}}}{2}\int\limits_{ – \pi }^\pi {\cos mxdx} }
+ {\sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }\right.}+{\left.{ {b_n}\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} } \right]} .}
\] The first term on the right side is zero. Then, using the well-known trigonometric identities, we have
\[
{\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} }
= {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\sin{\left( {n + m} \right)x} }\right.}+{\left.{ {\sin \left( {n – m} \right)x}} \right]dx} = 0,}
\] \[
{\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }
= {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\cos {\left( {n + m} \right)x} }\right.}+{\left.{ {\cos \left( {n – m} \right)x}} \right]dx} = 0,}
\] if \(m \ne n.\)

In case when \(m = n\), we can write:
\[\require{cancel}
{\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} }
= {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\sin 2mx + \sin 0} \right]dx} ,\;\;}\Rightarrow
{\int\limits_{ – \pi }^\pi {{\sin^2}mxdx} }={ \frac{1}{2}\left[ {\left. {\left( { – \frac{{\cos 2mx}}{{2m}}} \right)} \right|_{ – \pi }^\pi } \right] }
= {\frac{1}{{4m}}\left[ { – \cancel{\cos \left( {2m\pi } \right)} }\right.}+{\left.{ \cancel{\cos \left( {2m\left( { – \pi } \right)} \right)}} \right] }={ 0;}
\] \[
{\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }
= {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\cos 2mx + \cos 0} \right]dx} ,\;\;}\Rightarrow
{\int\limits_{ – \pi }^\pi {{\cos^2}mxdx} }
= {\frac{1}{2}\left[ {\left. {\left( {\frac{{\sin 2mx}}{{2m}}} \right)} \right|_{ – \pi }^\pi + 2\pi } \right] }
= {\frac{1}{{4m}}\left[ {\sin \left( {2m\pi } \right) }\right.}-{\left.{ \sin \left( {2m\left( { – \pi } \right)} \right)} \right] + \pi }={ \pi .}
\] Thus,
\[
{\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} = {a_m}\pi ,\;\;}\Rightarrow
{{a_m} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} ,\;\;}\kern-0.3pt
{m = 1,2,3, \ldots }
\] Similarly, multiplying the Fourier series by \(\sin mx\) and integrating term by term, we obtain the expression for \({{b_m}}:\)
\[
{{b_m} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin mxdx} ,\;\;\;}\kern-0.3pt
{m = 1,2,3, \ldots }
\] Rewriting the formulas for \({{a_n}},\) \({{b_n}},\) we can write the final expressions for the Fourier coefficients:
\[
{{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nxdx} ,\;\;\;}\kern-0.3pt
{{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nxdx} .}
\]

Example 2.

Find the Fourier series for the square \(2\pi\)-periodic wave defined on the interval \(\left[ { – \pi ,\pi } \right]:\)
\[
{f\left( x \right) \text{ = }}\kern0pt
{\begin{cases}
0, & \text{if} & – \pi \le x \le 0 \\
1, & \text{if} & 0 < x \le \pi
\end{cases}.}
\]

Solution.

First we calculate the constant \({{a_0}}:\)
\[
{{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} }
= {\frac{1}{\pi }\int\limits_0^\pi {1dx} }
= {\frac{1}{\pi } \cdot \pi }={ 1.}
\] Find now the Fourier coefficients for \(n \ne 0:\)
\[
{{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nxdx} }
= {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \cos nxdx} }
= {\frac{1}{\pi }\left[ {\left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^\pi } \right] }
= {\frac{1}{{\pi n}} \cdot 0 }={ 0,}
\] \[
{{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nxdx} }
= {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \sin nxdx} }
= {\frac{1}{\pi }\left[ {\left. {\left( { – \frac{{\cos nx}}{n}} \right)} \right|_0^\pi } \right] }
= { – \frac{1}{{\pi n}} \cdot \left( {\cos n\pi – \cos 0} \right) }
= {\frac{{1 – \cos n\pi }}{{\pi n}}.}
\] As \(\cos n\pi = {\left( { – 1} \right)^n},\) we can write:
\[{b_n} = \frac{{1 – {{\left( { – 1} \right)}^n}}}{{\pi n}}.\] Thus, the Fourier series for the square wave is
\[{f\left( x \right) = \frac{1}{2} }+{ \sum\limits_{n = 1}^\infty {\frac{{1 – {{\left( { – 1} \right)}^n}}}{{\pi n}}\sin nx} .}\] We can easily find the first few terms of the series. By setting, for example, \(n = 5,\) we get
\[
{f\left( x \right) = \frac{1}{2} }+{ \frac{{1 – \left( { – 1} \right)}}{\pi }\sin x }
+ {\frac{{1 – {{\left( { – 1} \right)}^2}}}{{2\pi }}\sin 2x }
+ {\frac{{1 – {{\left( { – 1} \right)}^3}}}{{3\pi }}\sin 3x }
+ {\frac{{1 – {{\left( { – 1} \right)}^4}}}{{4\pi }}\sin 4x }
+ {\frac{{1 – {{\left( { – 1} \right)}^5}}}{{5\pi }}\sin 5x + \ldots }
= {\frac{1}{2} + \frac{2}{\pi }\sin x }
+ {\frac{2}{{3\pi }}\sin 3x }
+ {\frac{2}{{5\pi }}\sin 5x + \ldots }
\]

Fourier series for the square  2pi-periodic wave

Figure 1, n = 10

The graph of the function and the Fourier series expansion for \(n = 10\) is shown in Figure \(1.\)

Page 1
Problems 1-2
Page 2
Problems 3-6